Đáp án đúng: B
10,53%.
Gọi số mol Fe và Fe3O4 trong mỗi phần lần lượt là x, y
Phần 1:
$\displaystyle \begin{array}{l}\text{F}{{\text{e}}_{\text{3}}}{{\text{O}}_{\text{4}}}\text{+ 8}{{\text{H}}^{\text{+}}}\xrightarrow{{}}\text{F}{{\text{e}}^{\text{2+}}}\text{+ 2F}{{\text{e}}^{\text{3+}}}\text{+ 4}{{\text{H}}_{\text{2}}}\text{O}\\\text{y }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\xrightarrow{{}}\text{ }\!\!~~~~\!\!\text{ 8y }\!\!~~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~~~~~~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\xrightarrow{{}}\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\,\,\,\,\,\,\,\text{2y}\\\text{Fe }\,\,\text{+ }\,\,\,\,\text{2}{{\text{H}}^{\text{+}}}\xrightarrow{{}}\text{F}{{\text{e}}^{\text{2+}}}\text{+ }{{\text{H}}_{\text{2}}}\\\text{x }\!\!~~~~\!\!\text{ }\xrightarrow{{}}\text{ }\!\!~\!\!\text{ 2x}\end{array}$
Ta có: 2x + y = 1,8 (1)
Phần 2:
$\displaystyle \begin{array}{l}\text{Fe }\xrightarrow{{}}\text{F}{{\text{e}}^{\text{3+}}}\text{+ 3e}\\\,\text{x}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\xrightarrow{{}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{3x}\\\text{F}{{\text{e}}_{\text{3}}}{{\text{O}}_{\text{4}}}\text{+ 8}{{\text{H}}^{\text{+}}}\text{ }\!\!~\!\!\text{ }\xrightarrow{{}}\text{3F}{{\text{e}}^{\text{3+}}}\text{ }\!\!~\!\!\text{ + }\!\!~\!\!\text{ }\,\text{4}{{\text{H}}_{\text{2}}}\text{O}\,\text{+ 1e}\\\text{y }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\xrightarrow{{}}\text{ }\!\!~\!\!\text{ 8y }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\xrightarrow{{}}\,\,\,\,\,\,\,\,\,\,\,\,\text{y}\\\text{S}{{\text{O}}_{\text{4}}}^{\text{2-}}\text{+ 4}{{\text{H}}^{\text{+}}}\text{+ 2e}\,\xrightarrow{{}}\,\text{S}{{\text{O}}_{\text{2}}}\text{+ 2}{{\text{H}}_{\text{2}}}\text{O}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{1,0 }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~~~~~\!\!\text{ 0,5 }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\,\,\,\,\,\,\,\,\,\,\text{0,25}\end{array}$
Áp dụng bảo toàn electron ta có: 3x + y = 0,5 (2)
Giải hệ (1) và (2) ta có: x = 0,1; y =0,2
Số mol H2SO4 đã phản ứng = ½ số mol H+ phản ứng = 1,3 mol
Khối lượng H2SO4 còn lại = Khối lượng H2SO4 ban đầu – Khối lượng H2SO4 phản ứng =147 – 1,3.98= 19,6 g
Khối lượng dung dịch sau phản ứng = 150 + khối lượng hỗn hợp – khối lượng tách ra (SO2)
= 150 + 0,1.56 + 0,2.232 – 0,25.64 = 186 gam
Nồng độ H2SO4 còn lại = 19,6.100/186 = 10,53%
